Trang chủ » Hỏi đáp » Môn Toán tìm x,y,z biết a, 5x=2y;3y=5z và x+y+z=-970 b,2x=-3y;7y=-10z và x-2y +3z=56 giúp vs ạ 18/11/2023 tìm x,y,z biết a, 5x=2y;3y=5z và x+y+z=-970 b,2x=-3y;7y=-10z và x-2y +3z=56 giúp vs ạ
Giải đáp: $\begin{array}{l}a)x = – 194;y = – 485;z = – 291\\b)\,x = 15;y = – 10;z = 7\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a)\left\{ \begin{array}{l}5x = 2y\\3y = 5z\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{x}{2} = \dfrac{y}{5}\\\dfrac{y}{5} = \dfrac{z}{3}\end{array} \right.\\ \Leftrightarrow \dfrac{x}{2} = \dfrac{y}{5} = \dfrac{z}{3} = \dfrac{{x + y + z}}{{2 + 5 + 3}} = \dfrac{{ – 970}}{{10}} = – 97\\ \Leftrightarrow \left\{ \begin{array}{l}x = – 97.2 = – 194\\y = – 97.5 = – 485\\z = – 97.3 = – 291\end{array} \right.\\Vay\,x = – 194;y = – 485;z = – 291\\b)\left\{ \begin{array}{l}2x = – 3y\\7y = – 10z\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{x}{{ – 3}} = \dfrac{y}{2} \Leftrightarrow \dfrac{x}{{15}} = \dfrac{y}{{ – 10}}\\\dfrac{y}{{ – 10}} = \dfrac{z}{7}\end{array} \right.\\ \Leftrightarrow \dfrac{x}{{15}} = \dfrac{y}{{ – 10}} = \dfrac{z}{7} = \dfrac{{2y}}{{ – 20}} = \dfrac{{3z}}{{21}}\\ = \dfrac{{x – 2y + 3z}}{{15 – \left( { – 20} \right) + 21}} = \dfrac{{56}}{{56}} = 1\\ \Leftrightarrow \left\{ \begin{array}{l}x = 1.15 = 15\\y = 1.\left( { – 10} \right) = – 10\\z = 1.7 = 7\end{array} \right.\\Vay\,x = 15;y = – 10;z = 7\end{array}$ Trả lời
a)x = – 194;y = – 485;z = – 291\\
b)\,x = 15;y = – 10;z = 7
\end{array}$
a)\left\{ \begin{array}{l}
5x = 2y\\
3y = 5z
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{2} = \dfrac{y}{5}\\
\dfrac{y}{5} = \dfrac{z}{3}
\end{array} \right.\\
\Leftrightarrow \dfrac{x}{2} = \dfrac{y}{5} = \dfrac{z}{3} = \dfrac{{x + y + z}}{{2 + 5 + 3}} = \dfrac{{ – 970}}{{10}} = – 97\\
\Leftrightarrow \left\{ \begin{array}{l}
x = – 97.2 = – 194\\
y = – 97.5 = – 485\\
z = – 97.3 = – 291
\end{array} \right.\\
Vay\,x = – 194;y = – 485;z = – 291\\
b)\left\{ \begin{array}{l}
2x = – 3y\\
7y = – 10z
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{{ – 3}} = \dfrac{y}{2} \Leftrightarrow \dfrac{x}{{15}} = \dfrac{y}{{ – 10}}\\
\dfrac{y}{{ – 10}} = \dfrac{z}{7}
\end{array} \right.\\
\Leftrightarrow \dfrac{x}{{15}} = \dfrac{y}{{ – 10}} = \dfrac{z}{7} = \dfrac{{2y}}{{ – 20}} = \dfrac{{3z}}{{21}}\\
= \dfrac{{x – 2y + 3z}}{{15 – \left( { – 20} \right) + 21}} = \dfrac{{56}}{{56}} = 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1.15 = 15\\
y = 1.\left( { – 10} \right) = – 10\\
z = 1.7 = 7
\end{array} \right.\\
Vay\,x = 15;y = – 10;z = 7
\end{array}$